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A bakery sells 4 varieties of pastries. John buys 10 pastries making

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A bakery sells 4 varieties of pastries. John buys 10 pastries making  [#permalink]

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A bakery sells 4 varieties of pastries. John buys 10 pastries making sure, he gets at least one of each kind. In how many ways can John make the purchase ?

Can anyone let me know how to solve this problem using the most efficient method ? What would be the answer to this word problem?
I don't know the answer and i am getting different values using different method.

Thanks
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Re: A bakery sells 4 varieties of pastries. John buys 10 pastries making  [#permalink]

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New post 03 Jul 2016, 07:49
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no of 1st variety of pastry: a
no of 2nd variety of pastry: b
no of 3rd variety of pastry: c
no of 4th variety of pastry: d

So, we have a+b+c+d=10 (where a,b,c,d>0)

Remember: Number of positive integer solution of x+y+z+... (r terms) = n is (n-1)C(r-1)

So, the answer to the question would be (10-1)C(4-1) = 9C3 = 84.

------------------------------------

P.S. Don't forget to give kudos ;)
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Re: A bakery sells 4 varieties of pastries. John buys 10 pastries making  [#permalink]

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New post Updated on: 03 Jul 2016, 22:52
14101992 wrote:
no of 1st variety of pastry: a
no of 2nd variety of pastry: b
no of 3rd variety of pastry: c
no of 4th variety of pastry: d

So, we have a+b+c+d=10 (where a,b,c,d>0)

Remember: Number of positive integer solution of x+y+z+... (r terms) = n is (n-1)C(r-1)

So, the answer to the question would be (10-1)C(4-1) = 9C3 = 84.

------------------------------------

P.S. Don't forget to give kudos ;)


Ok, your explanation sounds perfect.

My approach was:-
In a non restrictive way when a person buys 10 pastries but it is not compulsory to buy all four pastries:-apple+banana+cherry+dark chocolate =10 (four types of pastries should add up to 10)
Now a person can choose to ignore a or b or c or d completely
In that case the a or b or c or d can be zero
But since John need to buy atleast one pastry of each type therefore a,b,c,d cannot be 0
Therefore:- IN JOHN's CASE
(1+a)+(1+b)+(1+c)+(1+d)=10
a+b+c+d=10-4
a+b+c+d=6
Then I used P&C partition method
3 partitions needed to make 6
| | | | | | P P P (Total object to chose from= 9 ; 6 of them are | and 3 of them are P)

therefore total unique way = nCr = 9C3 =\(\frac{9!}{3!6!}\)
==>\(\frac{9*8*7}{3*2*1}\)

==> \(\frac{504}{6}= 84\)

Is there a derivative proof for the easy method you used
Number of positive integer solution of x+y+z+... (r terms) = n
is (n-1)C(r-1)

_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

Originally posted by LogicGuru1 on 03 Jul 2016, 12:43.
Last edited by LogicGuru1 on 03 Jul 2016, 22:52, edited 2 times in total.
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Re: A bakery sells 4 varieties of pastries. John buys 10 pastries making  [#permalink]

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New post 03 Jul 2016, 20:42
LogicGuru1 wrote:
14101992 wrote:
no of 1st variety of pastry: a
no of 2nd variety of pastry: b
no of 3rd variety of pastry: c
no of 4th variety of pastry: d

So, we have a+b+c+d=10 (where a,b,c,d>0)

Remember: Number of positive integer solution of x+y+z+... (r terms) = n is (n-1)C(r-1)

So, the answer to the question would be (10-1)C(4-1) = 9C3 = 84.

------------------------------------

P.S. Don't forget to give kudos ;)


Ok, your explanation sounds perfect.

My approach was:-
In a non restrictive way when a person buys 10 pastries but it is not compulsory to buy all four pastries:-apple+banana+cherry+dark chocolate =10 (four types of pastries should add up to 10)
Now a person can choose to ignore a or b or c or d completely
In that case the a or b or c or d can be zero
But since John need to buy atleast one pastry of each type therefore neither a,b,c,d can be 0
Therefore:- IN JOHN's CASE
(1+a)+(1+b)+(1+c)+(1+d)=10
a+b+c+d=10-4
a+b+c+d=6
Then I used P&N partition method
3 partitions needed to make 6
| | | | | | P P P (Total object to chose from= 9 ; 6 of them are | and 3 of them are P)

therefore total unique way = nCr = 9C3 =\(\frac{9!}{3!6!}\)
==>\(\frac{9*8*7}{3*2*1}\)

==> \(\frac{504}{6}= 84\)

Is there a derivative proof for the easy method you used
Number of positive integer solution of x+y+z+... (r terms) = n
is (n-1)C(r-1)


The formula is doing exactly what you are doing.

You need 10 goodies. You need to split them into 4 distinct groups to decide how many of each should you take.

G _ G _ G _ G _ G _ G _ G _ G _ G _ G

There are 9 slots to put in the partition. (shown by _ ). We need to select any 3 of them since we need to make 4 groups. Each group will have at least one goodie.
That is how you get 9C3 or in the general case (n-1)C(r-1)
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Re: A bakery sells 4 varieties of pastries. John buys 10 pastries making  [#permalink]

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New post 03 Jul 2016, 22:19
LogicGuru1 wrote:
14101992 wrote:
no of 1st variety of pastry: a
no of 2nd variety of pastry: b
no of 3rd variety of pastry: c
no of 4th variety of pastry: d

So, we have a+b+c+d=10 (where a,b,c,d>0)

Remember: Number of positive integer solution of x+y+z+... (r terms) = n is (n-1)C(r-1)

So, the answer to the question would be (10-1)C(4-1) = 9C3 = 84.

------------------------------------

P.S. Don't forget to give kudos ;)


Ok, your explanation sounds perfect.

My approach was:-
In a non restrictive way when a person buys 10 pastries but it is not compulsory to buy all four pastries:-apple+banana+cherry+dark chocolate =10 (four types of pastries should add up to 10)
Now a person can choose to ignore a or b or c or d completely
In that case the a or b or c or d can be zero
But since John need to buy atleast one pastry of each type therefore neither a,b,c,d can be 0
Therefore:- IN JOHN's CASE
(1+a)+(1+b)+(1+c)+(1+d)=10
a+b+c+d=10-4
a+b+c+d=6
Then I used P&N partition method
3 partitions needed to make 6
| | | | | | P P P (Total object to chose from= 9 ; 6 of them are | and 3 of them are P)

therefore total unique way = nCr = 9C3 =\(\frac{9!}{3!6!}\)
==>\(\frac{9*8*7}{3*2*1}\)

==> \(\frac{504}{6}= 84\)

Is there a derivative proof for the easy method you used
Number of positive integer solution of x+y+z+... (r terms) = n
is (n-1)C(r-1)


On the same lines, think about what happens when you are allowed to take no pastry of one or more types, that is, when the condition of "at least one pastry of each type" is dropped.
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Re: A bakery sells 4 varieties of pastries. John buys 10 pastries making  [#permalink]

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New post 03 Jul 2016, 23:00
VeritasPrepKarishma wrote:

On the same lines, think about what happens when you are allowed to take no pastry of one or more types, that is, when the condition of "at least one pastry of each type" is dropped.


Well i guess the Natural number solution (1 to 10 in this case) will change to a whole number solution 0 to 10 (Why don' people use these terms here, i don't understand, I know gmat is obsessed with using INTEGER as a larger subset.. But using terms like natural number, counting numbers and whole numbers once in a while is a good reason to fire the lazy neurons :))

so coming back to the point ..

The solution will be
\(\frac{13!}{10!3!}\)
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
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Re: A bakery sells 4 varieties of pastries. John buys 10 pastries making   [#permalink] 03 Jul 2016, 23:00

A bakery sells 4 varieties of pastries. John buys 10 pastries making

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